1007 DNA Sorting

Description

One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC'', this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG'' has only one inversion (E and D)---it is nearly sorted---while the sequence ``ZWQM'' has 6 inversions (it is as unsorted as can be---exactly the reverse of sorted). 

You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness'', from ``most sorted'' to ``least sorted''. All the strings are of the same length. 

Input

The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (0 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.

Output

Output the list of input strings, arranged from ``most sorted'' to ``least sorted''. Since two strings can be equally sorted, then output them according to the orginal order.

Sample Input

10 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

CODE:

#include <iostream>using namespace std;typedef struct  {char* val;int key;} node;int getnum(char* s){int sum=0;char *s2,*s3;for(s2=s;*s2!='\0';++s2)for(s3=s2+1;*s3!='\0';++s3)if(*s2-*s3>0)++sum;return sum;}int main(int argc, char* argv[])  {  int m,n,i;cin>>m>>n;node *in=new node[n];node temp;for(i=0;i<n;++i){in[i].val=new char[m+1];cin>>in[i].val;in[i].key=getnum(in[i].val);}bool moved=true;i=0;while(moved){moved=false;for(i=0;i<n-1;++i)if(in[i].key>in[i+1].key){moved=true;temp=in[i];in[i]=in[i+1];in[i+1]=temp;}}for(i=0;i<n;++i){cout<<in[i].val<<endl;delete in[i].val;}delete in;return 0;}